t2+4t+19=42

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Solution for t2+4t+19=42 equation:



t2+4t+19=42
We move all terms to the left:
t2+4t+19-(42)=0
We add all the numbers together, and all the variables
t^2+4t-23=0
a = 1; b = 4; c = -23;
Δ = b2-4ac
Δ = 42-4·1·(-23)
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{3}}{2*1}=\frac{-4-6\sqrt{3}}{2} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{3}}{2*1}=\frac{-4+6\sqrt{3}}{2} $

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